Step 1: Write expressions for the mean and variance.
For a binomial distribution $B(n, p)$, the mean is $\mu = np$ and the variance is $\sigma^2 = npq$, where $q = 1 - p$.
Step 2: State the condition given in the question.
'Success is not an impossible event' means $p > 0$. Since $p$ is also at most 1, we have $0 < p \leq 1$, which gives $0 \leq q < 1$.
Step 3: Compare mean and variance symbolically.
We compute $\mu - \sigma^2 = np - npq = np(1 - q) = np \cdot p = np^2$.
Step 4: Determine the sign of the difference.
Since $n \geq 1$, $p > 0$, we have $np^2 > 0$. Therefore $\mu - \sigma^2 > 0$, i.e. $\mu > \sigma^2$. The mean is strictly greater than the variance.
Step 5: Check the extreme case $p = 1$.
If $p = 1$, then $q = 0$ and $\sigma^2 = n \cdot 1 \cdot 0 = 0$, while $\mu = n > 0$. Mean is still greater than variance.
Step 6: Eliminate the other options.
The mean equals the variance only when $p = 0$ (success is impossible -- excluded by the condition). The mean is less than the variance only when $q > 1$, which is impossible for a probability.
Step 7: State the final answer.
The mean of a binomial distribution is \[ \boxed{\text{always more than its variance}} \]